A) Antiparallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\]
B) Zero
C) Parallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\]
D) Parallel to \[\hat{k}\]
Correct Answer: A
Solution :
| [a] \[\vec{E}\] at t = 0 at \[z=\pi k\] is given by |
| \[\vec{E}=\frac{{{E}_{0}}}{\sqrt{2}}\,(\hat{i}+\hat{j})\,cos\,[\pi ]\,=-\,\frac{{{E}_{0}}}{\sqrt{2}}\,(\hat{i}+\hat{j})\] |
| \[{{\vec{F}}_{E}}=q\vec{E}\] |
| \[\Rightarrow \] Force due to electric field, \[{{\vec{F}}_{E}}\] is antiparallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\]. |
| \[{{\vec{F}}_{mag}}=q(\vec{v}\times \vec{B})\] |
| \[\vec{B}\,\,(at\,t=0,\,z\,=\pi k)\]is \[\frac{{{B}_{0}}}{\sqrt{2}}\,(-\hat{i}+\hat{j})\] |
| \[\Rightarrow \,\,\,{{\vec{F}}_{mag}}\,=q\,{{v}_{0}}\hat{k}\times \frac{{{B}_{0}}}{\sqrt{2}}\,(-\hat{i}+\hat{j})\]which is antiparallel |
| to \[\frac{(\hat{i}+\hat{j})}{\sqrt{2}}\] |
| \[\Rightarrow \,\,{{\vec{F}}_{net}}={{\vec{F}}_{E}}+{{\vec{F}}_{B}}\] is Antiparallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\] |
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